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Dec 27, 2021 at 15:08 comment added Kensuke_Yamato I think I got it. After taking resolution, we could see that any exceptional divisor $E$ for $X$ on $\tilde{X}$ satisfies $\Phi_* \mathcal{O}_{\tilde{X}}(E) \simeq \mathcal{O}_Y$ as in your answer for [2]. Thank you for answering.
Dec 27, 2021 at 14:34 vote accept Kensuke_Yamato
Dec 27, 2021 at 14:32 vote accept Kensuke_Yamato
Dec 27, 2021 at 14:34
Dec 26, 2021 at 18:29 history answered Hacon CC BY-SA 4.0